Question

Evaluate the definite integral: \( \displaystyle \int_{3}^{5} |x-4|\,dx \).

• A. \(1\)
• B. \(2\)
• C. \(3\)
• D. \(4\)

Hint:

When integrating absolute value functions, always split the interval at the point where the expression inside the modulus becomes zero.

Answer 1

The Correct Option is B

Concept: For absolute value functions, \[ |x-a|= \begin{cases} x-a & x \ge a \\ a-x & x<a \end{cases} \] Thus the integral must be split at the point where the expression inside the modulus becomes zero.

Step 1: Find where the expression inside modulus becomes zero. \[ x-4=0 \quad \Rightarrow \quad x=4 \] So we split the integral: \[ \int_{3}^{5}|x-4|dx = \int_{3}^{4}(4-x)\,dx + \int_{4}^{5}(x-4)\,dx \]

Step 2: Evaluate the first integral. \[ \int_{3}^{4}(4-x)\,dx = \left[4x-\frac{x^2}{2}\right]_{3}^{4} \] \[ = (16-8)-(12-\tfrac{9}{2}) \] \[ =8-\tfrac{15}{2} =\tfrac{1}{2} \]

Step 3: Evaluate the second integral. \[ \int_{4}^{5}(x-4)\,dx = \left[\frac{x^2}{2}-4x\right]_{4}^{5} \] \[ = \left(\tfrac{25}{2}-20\right)-(8-16) \] \[ =\tfrac{1}{2} \]

Step 4: Add the results. \[ \tfrac{1}{2}+\tfrac{1}{2}=1 \] Thus, \[ \boxed{\int_{3}^{5}|x-4|dx=1} \]